CodeForces-148A-Insomnia cure

公主在夜晚想象龙偷走她,并通过不同的惩罚来应对每只龙,计算实际受到伤害的龙的数量。

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A. Insomnia cure
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.

However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every k-th dragon got punched in the face with a frying pan. Every l-th dragon got his tail shut into the balcony door. Every m-th dragon got his paws trampled with sharp heels. Finally, she threatened every n-th dragon to call her mom, and he withdrew in panic.

How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of d dragons?

Input

Input data contains integer numbers k, l, m, n and d, each number in a separate line (1 ≤ k, l, m, n ≤ 101 ≤ d ≤ 105).

Output

Output the number of damaged dragons.

Sample test(s)
input
1
2
3
4
12
output
12
input
2
3
4
5
24
output
17
Note

In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.

In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.


import java.util.*;
public class InsomniaCure {
	public static void main(String[]args){
		Scanner in=new Scanner(System.in);
		int k=in.nextInt();
		int l=in.nextInt();
		int m=in.nextInt();
		int n=in.nextInt();
		int d=in.nextInt();
		int count=0;
		for(int i=1;i<=d;i++){
			if(i%k!=0&&i%l!=0&&i%m!=0&&i%n!=0)
				count++;
		}
		System.out.println(d-count);
	}
}

tips:1、<span style="font-family: Arial, Helvetica, sans-serif;">在k l m n 这几个数据的录入中可以抽象为数组内四位元素,从而采用循环赋值方式,更加方便 。2、if判断句可用德摩根律转换成并集形式</span>
### 关于 Codeforces Problem 1802A 目前提供的引用内容并未涉及 Codeforces 编号为 1802A 的题目详情或解决方案[^1]。然而,基于常见的竞赛编程问题模式以及可能的解决方法,可以推测该类题目通常围绕算法设计、数据结构应用或者特定技巧展开。 如果假设此题属于典型的算法挑战之一,则可以从以下几个方面入手分析: #### 可能的方向一:字符串处理 许多入门级到中级难度的问题会考察字符串操作能力。例如判断子串是否存在、统计字符频率或是执行某种转换逻辑等。以下是 Python 中实现的一个简单例子用于演示如何高效地比较两个字符串是否相匹配: ```python def are_strings_equal(s1, s2): if len(s1) != len(s2): return False for i in range(len(s1)): if s1[i] != s2[i]: return False return True ``` #### 方向二:数组与列表的操作 另一常见主题是对整数序列进行各种形式上的变换或者是查询最值等问题。下面给出一段 C++ 程序片段来展示快速寻找最大元素位置的方法: ```cpp #include <bits/stdc++.h> using namespace std; int main(){ int n; cin >> n; vector<int> a(n); for(auto &x : a){ cin>>x; } auto max_it = max_element(a.begin(),a.end()); cout << distance(a.begin(),max_it)+1; // 输出索引加一作为答案 } ``` 由于具体描述缺失,在这里仅提供通用框架供参考。对于确切解答还需要访问实际页面获取更多信息后再做进一步探讨[^3]。
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